State-space introduction
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| state space | không gian trạng thái |
| eigenvalue/ˈaɪɡənvæljuː/ | giá trị riêng |
| transition matrix/trænˈsɪʃn ˈmeɪtrɪks/ | ma trận chuyển trạng thái |
| controllability | khả năng điều khiển |
n drawers of memory
- State space 状态空间 writes $\dot x = Ax + Bu$, $y = Cx + Du$: the states are $n$ drawers of memory, $A$ says how they push each other, $B$ where the input enters, $C$ what you measure.
- The eigenvalues 特征值 of $A$ are the poles, because $\det(sI - A) = 0$ is the characteristic equation. An eigenvalue is a number lambda for which the state matrix has a nonzero solution direction, and its real part sets how fast that direction decays or grows.
$G = \dfrac{8}{(s+2)(s+4)}$ in companion form, with the DC-gain check $-CA^{-1}B = 8/8 = 1$ and eigenvalues $-2, -4$: the poles.
{exercise:uni-ee1048-form}
For G = 8/[(s+2)(s+4)] in companion form with A = [[0,1],[−8,−6]], B = [0;1], C = [8,0], the DC gain −CA⁻¹B equals —
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
−CA⁻¹B = 8/8 = 1, matching G(0) = 8/(2×4). The DC-gain check catches a misplaced C or sign in one line.
The transition matrix is the poles made real
- With zero input, $x(t) = \Phi(t)\,x(0)$ where the transition matrix 状态转移矩阵 $\Phi(t) = e^{At} = I + At + \frac{(At)^2}{2!} + \cdots$.
- Every eigenvalue $\lambda$ of $A$ becomes the decay factor $e^{\lambda t}$ in $\Phi$: the matrix power is the pole.
![State trajectories of the model of 8/[(s+2)(s+4)] in the phase plane](/handout-media/university_electrical_engineering/assets/10-state-portrait.png?v=1793069544)
With eigenvalues $e^{-1} = 0.368$ and $e^{-2} = 0.135$. From $x(0) = [1, 0]^T$ the state after 0.5 s is the first column, decaying along the slow eigenvector $[1, -2]^T$.
{exercise:uni-ee1048-phi}
Φ(0.5) = e^{0.5A} for that model has eigenvalues —
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
Each pole λ of A becomes e^{λt} in the transition matrix: at t = 0.5 that is e^−1 and e^−2. The matrix power is the pole, made real.
Pole placement by linear algebra
- Full-state feedback $u = -K_{fb}x$ (a gain row) gives the closed matrix $A - BK_{fb}$: match its characteristic polynomial to the one you want and solve for the gains.
- Demand poles at $-10, -10$ for our model: $s^{2}+(6+k_2)s+(8+k_1) = s^{2}
- 20s + 100$gives$K_{fb} = [92, 14]$, and the eigenvalues of$A - BK_{fb}$compute to$-10, -10$ exactly.
The guarantee is controllability 能控性: the rank of $[\,B\;\;AB\;\cdots\,]$ must be $n$ (rank 2 here). Controllability is the guarantee that some input sequence can drive the state to any target in finite time. What locus search does along a ray, state feedback does by solving linear equations.
{exercise:uni-ee1048-place}
A = [[0,1],[−10,−7]], B = [0;1]. Placing both poles at −8 and −12 needs u = −Kx with K = —
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
A − BK has char s² + (7+k₂)s + (10+k₁); matching s² + 20s + 96 gives k₁ = 86, k₂ = 13. Check: eigenvalues of A − BK are exactly −8 and −12.
Two conventions to keep straight
- The output is $Cx$, not the states: the transfer function is $C(sI-A)^{-1}B + D$, and forgetting $C$ changes the DC gain.
- $e^{At}$ is defined by the series: its off-diagonal entries are not exponentials of the entries. Row-state textbooks flip every product relative to this column convention.
{exercise:uni-ee1048-conv}
The transfer function of the state-space model (D = 0) is —
Độ khó 1/5 · 100 xu + phần thưởng nhỏ khi hoàn thành lần đầu tiên hôm nay
Transfer function = C(sI−A)⁻¹B + D. Dropping C returns a state-to-state object and quietly changes the DC gain; the order C, (sI−A)⁻¹, B is fixed by column states.